Given the number k, return the minimum number of Fibonacci numbers whose sum is equal to k, whether a Fibonacci number could be used multiple times. The Fibonacci numbers are defined as:

F1 = 1

F2 = 1

Fn=Fn-1 + Fn-2, for n > 2.

It is guaranteed that for the given constraints we can always find such Fibonacci numbers that sum k.

Input: k=7

Output: 2

def fibo(x): if x <= 1: return x else: return (fibo(x-1)+fibo(x-2)) N=int(input('Enter the K value : ')) L=[] for i in range(N): if fibo(i) < N: L.append(fibo(i)) else: break i,c=0,0 L=sorted(L,reverse=True) while i<len(L)-1: c+=N//L[i] N%=L[i] i+=1 print(c)

Input_1:

Enter the K value : 7

Output:

2

Input_2:

Enter the K value : 10

Output:

2

Input_3:

Enter the K value : 19

Output:

3

Input_4:

Enter the K value : 900

Output:

4

Input_5:

Enter the K value : 17

Output:

3

**Morae Q!**

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